evaluate the series 4 sigma n=1 n+4

Answer: [tex]\sum _{n=1}^4n+4=26[/tex]Step-by-step explanation:  Given : [tex]\sum _{n=1}^4\:n+4[/tex]We have to evaluate the sum.  Consider , the given sum [tex]\sum _{n=1}^4\:n+4[/tex]Apply the sum rule, [tex]\sum a_n+b_n=\sum a_n+\sum b_n[/tex]    we have,[tex]=\sum _{n=1}^4n+\sum _{n=1}^44[/tex]Consider [tex]\sum _{n=1}^4n[/tex] first,Applying sum formula, [tex]\sum _{k=1}^nk=\frac{1}{2}n\left(n+1\right)[/tex] Here, n = 4, we get,[tex]=\frac{1}{2}\cdot \:4\left(4+1\right)[/tex][tex]\sum _{n=1}^4n=10[/tex]Now consider [tex]\sum _{n=1}^44[/tex][tex]\mathrm{Apply\:the\:Sum\:Formula:\quad }\sum _{k=1}^n\:a\:=\:a\cdot n[/tex][tex]=4\cdot \:4=16[/tex]Therefore, [tex]=\sum _{n=1}^4n+\sum _{n=1}^44=10+16=26[/tex]Thus, [tex]\sum _{n=1}^4n+4=26[/tex]